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Choose the correct option from the given alternatives : If andthendydxxy+1+yx+1=0andx≠y,thendydx = ........ - Mathematics and Statistics

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प्रश्न

Choose the correct option from the given alternatives :

If `xsqrt(y + 1) + ysqrt(x + 1) = 0 and x ≠ y, "then" "dy"/"dx"` = ........

विकल्प

  • `(1)/(1 + x)^2`

  • `-(1)/(1 + x)^2`

  • (1 + x)2 

  • `-x/(x + 1)`

MCQ
योग
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उत्तर

`-(1)/(1 + x)^2`

Explanation:

`xsqrt(y + 1) = -ysqrt(x + 1)`

Squaring both the sides,

∴ x2(y + 1) = y2(x + 1)

∴ x2y + x2 = xy2 + y2

∴ x2 – y2 = xy2 – x2y

∴ (x – y)(x + y) = – xy(x – y)

∴ x + y = – xy                      ...[∵ x ≠ y] 

∴ x = – xy – y

∴ x = – y (x + 1)

∴ y = `- x/(x + 1)`

Differentiating both sides w.r.t.x, we get

`dy/dx = - [(1 + x) d/dx(x) - (x) d/dx (x + 1)]/(1 + x)^2`

`dy/dx = - [(1 + x). 1 - x(1 + 0)]/(1 + x)^2`

`dy/dx = - [1 + cancelx - cancelx]/(1 + x)^2`

∴ `"dy"/"dx" = -(1)/(1 + x)^2`.

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अध्याय 1: Differentiation - Miscellaneous Exercise 1 (I) [पृष्ठ ६२]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Miscellaneous Exercise 1 (I) | Q 9 | पृष्ठ ६२

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